Let $K$ be an algebraic number field and $\mathcal{O}_K$ the integral closure of $\Bbb{Z}$ in $K$. Let us recall that a Dedekind domain is a UFD iff it is a PID. Classically, I think that the questions concerning whether or not a certain Dedekind domain was a UFD were very important, see e.g. this thread here.
Perhaps from the point of view of algebra asking if something is a PID is easier to approach: we know how to factor ideals in Dedekind domains and thus there should at least be a tool to measure how far does a Dedekind domain differ from being a principal ideal domain.
Let me now give you an alternative definition of the ideal class group. We will put an equivalence relation on the set of all ideals defined as follows. We say that an ideal $I$ of $\mathcal{O}_K$ is equivalent to $J$ iff there is $\alpha,\beta \in \mathcal{O}_K$ so that
$$\alpha I = \beta J.$$
One easily checks that this is an equivalence relation. With a little bit more work, one can show that the set of all equivalence classes has a well defined multiplication law and is actually a group. The identity element being the class of all principal ideals. Now for some exercises.
Exercise 1: Check that the "class" of all principal ideals is actually a class. Namely if $I$ is an ideal such that $\alpha I = (\beta)$ then show that $I$ is actually principal. Hint: $\mathcal{O}_K$ is an integral domain.
Exercise 2: Show that this definition of an ideal class group is actually equivalent to the one given in Neukirch. Hint: Use the first isomorphism theorem.
Now do you see how the definition I have given you of an ideal class group actually measures nicely whether or not $\mathcal{O}_K$ is a PID? We see that $\mathcal{O}_K$ is a PID iff its ideal class group is trivial.
Now on to more interesting material. In advanced subjects such as Class Field Theory one can construct something know as the Hilbert Class Field of $K$. I don't know all the details of this construction as my algebraic number theory is not so advanced, but in the Hilbert class field every ideal of $\mathcal{O}_K$ becomes principal!! One can now ask the question: can we avoid talking of the Hilbert class field and find such an extension?
The answer is: Of course we can! This is where the ideal class group comes in. Firstly from Minkowski's bound we get that $Cl_K$ is actually a finite group. Using this, here are now two exercises which you can do:
Exercise 3: Let $I$ be an ideal of $\mathcal{O}_K$ show that there is a finite extension $L/K$ so that $I\mathcal{O}_L$ is principal. Hint: By finiteness of the class group there is an $n$ so that $I^n = \alpha$ for some $\alpha \in \mathcal{O}_K$. Now consider $L = K(\sqrt[n]{\alpha})$.
Exercise 4: Show that there is a finite extension of $L$ in which every ideal of $\mathcal{O}_K$ becomes principal. Wowowowowow!
Hint: Use exercise 3 and the definition of the ideal class group given in the beginning of my answer, not the one in Neukirch.
If you are stuck with any of these exercises I can post their solutions for you to view here.
Solution to Exercise 1 (As requested by user Andrew):
Suppose that $\alpha I = (\beta)$. Then in particular there is $x \in I$ so that $\alpha x = \beta$. We claim that $I = (x)$. Now it is clear that $(x) \subseteq I$. For the reverse inclusion take any $y \in I$. Then $\alpha y = \beta \gamma$ for some $\gamma \in R$. Since $\beta = \alpha x$ we get that
$$\alpha y = \alpha x \gamma.$$
But now $\mathcal{O}_K$ is an integral domain and so $y = x\gamma$, so that $I \subseteq (x)$. Hence $I = (x)$ and so $I$ is principal.
Best Answer
Every element of $K^{\times}$ generates a principal fractional ideal, and two elements of $K^{\times}$ generate the same principal fractional ideal if they differ by a unit of $\mathcal{O}_K$, i.e. by an element of $\mathcal{O}^{\times}_K$. In this sense the unit group $\mathcal{O}_K^{\times}$ measures the proportion of numbers to principal fractional ideals. Here numbers is understood to mean elements of $K^{\times}$.
This characterization also shows that the group $P_K$ of principal fractional ideals of $K$ is precisely the cokernel of the map $$1\ \longrightarrow\ \mathcal{O}_K^{\times}\ \longrightarrow\ K^{\times}.\tag{1}$$ That is to say, the group $P_K$ fits into the short exact sequence $$1\ \longrightarrow\ \mathcal{O}_K^{\times}\ \longrightarrow\ K^{\times}\ \longrightarrow\ P_K\ \longrightarrow\ 1.\tag{2}$$
Of course not all fractional ideals are principal in general. That is to say, going from numbers to ideals we also gain new ideals, apart from those generated by single numbers (i.e. principal ideals). The proportion of fractional ideals to principal fractional ideals is measured by the quotient $J_K/P_K$. This quotient is called the class group of $K$, denoted by $\operatorname{Cl}_K$.
This can be phrased briefly by saying that the class group $\operatorname{Cl}_K$ of $K$ is the cokernel of the map $$1\ \longrightarrow\ P_K\ \longrightarrow\ J_K.\tag{3}$$ Then by definition $\operatorname{Cl}_K$ fits into the short exact sequence $$1\ \longrightarrow\ P_K\ \longrightarrow\ J_K\ \longrightarrow\ \operatorname{Cl}_K\ \longrightarrow\ 1.\tag{4}$$ Putting the two short exact sequence $(2)$ and $(4)$ together shows that $\operatorname{Cl}_K$ fits into the exact sequence $$1\ \longrightarrow\ \mathcal{O}_K^{\times}\ \longrightarrow\ K^{\times}\ \longrightarrow\ J_K\ \longrightarrow\ \operatorname{Cl}_K\ \longrightarrow\ 1.$$ This shows that the group of units and the class group are the kernel and cokernel, respectively, of the map $K^{\times}\ \longrightarrow\ J_K$. The former measures how many numbers contract to the same ideal, the latter measures the proportion of ideals coming from numbers.